Closed form for an almost-elliptic integral

Does

$$\int_0^{2 \pi} \log\left(\frac{1}{2}[1+\sqrt{1-(a \sin\phi)^2}]\right) d\phi$$ have a closed form ?

An approximation for small $a$ is $2E-\pi$, but it is the exact form that is needed for any $|a|<1$.

The integration standing to Jack is

$$-\frac{\pi a^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;a^2\right)$$

The question now is: Can it be changed into a finite combination of elementary functions.

• Are you sure about $2e-\pi$ for small $a$ ? This does not match what I obtain using numerical integration. Could you check if the formula is correct ?( – Claude Leibovici Mar 27 '15 at 10:35
• Yes, it is E-pi. There is a missing parenthesis as mentionned by David. It is the integral from 0 to 2*pi of ln[0.5*(1+sqrt(1-square(a*sin(phi)))) ]. – MOEL Mar 27 '15 at 10:48
• Sorry again, 2E-pi not E -pi. – MOEL Mar 27 '15 at 10:49
• Is it possible to make the correction adding the missing parnethesis? – MOEL Mar 27 '15 at 10:51
• Jack, E is not the euler number but the elliptic E(square(a)). – MOEL Mar 27 '15 at 10:54

For any $b\in(0,1)$ we have: $$I(b)=\int_{0}^{2\pi}\log\left(1+\sqrt{1-b^2\sin^2 x}\right)\,dx=4\int_{0}^{1}\frac{\log\left(1+\sqrt{1-b^2 t^2}\right)}{\sqrt{1-t^2}}\,dt$$ hence: $$I(b)=\frac{1}{b}\int_{0}^{b}\frac{\log(1+\sqrt{1-u^2})}{\sqrt{1-\frac{u^2}{b^2}}}\,du=\frac{1}{b}\int_{0}^{\arcsin b}\frac{\cos\theta}{\sqrt{1-\frac{\sin^2\theta}{b^2}}}\,\log(1+\cos\theta)\,d\theta$$ and the last integral can be evaluated by exploiting the Fourier series of $\log(1+\cos\theta)$ that is pretty well-known. Another possible approach is differentiation under the integral sign.
We have: $$I'(b) = -4\int_{0}^{\pi/2}\frac{b\sin^2 t}{\sqrt{1-b^2\sin^2 t}\left(1+\sqrt{1-b^2\sin^2 t}\right)}\,dt = \frac{2\pi-4K(b^2)}{b}\tag{1}$$ where $K$ is the complete elliptic integral of the first kind. Since $I(0)=2\pi\log 2$, it follows that:
$$I(b)=2\pi\log 2-\frac{\pi b^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;b^2\right).$$
• @MOEL: that does not change anything. $\log(z/2)$ is just $\log z-\log 2$, hence get rid of the $-\log 2$ and take $b=a$. – Jack D'Aurizio Mar 27 '15 at 11:21